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When 6 boys were admitted and 6 girls left, the percentage of boys increased from $60\%$ to $75\%$. Find the original number of boys and girls in the class.

Last updated date: 17th Jul 2024
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Hint: According to the given question we are provided with some information about boys and girls in the class and we need to find the number of boys and girls using the given information. Also, we know that after the given condition the percentage of boys increased from $60\%$ to $75\%$.

According to the given question and information of the students we are being asked to find the number of girls and boys in the class. Now let us suppose that the number of boys is x and the number of girls is y. So, from this we get that the total students of the class are $x+y$ .
Now, we are given that initially boys percent was $60\%$which can be written as $\dfrac{x}{x+y}=\dfrac{60}{100}$ and when 6 boys were admitted and 6 girls left the percent becomes $75\%$which can be written as $\dfrac{x+6}{\left( x+6 \right)+\left( y-6 \right)}=\dfrac{75}{100}$ .
\begin{align} & \dfrac{x}{x+y}=\dfrac{60}{100} \\ & \Rightarrow 100x=60x+60y \\ & \Rightarrow 40x=60y \\ & \Rightarrow 2x=3y \\ \end{align}
\begin{align} & \dfrac{x+6}{\left( x+6 \right)+\left( y-6 \right)}=\dfrac{75}{100} \\ & \Rightarrow 100x+600=75x+75y \\ & \Rightarrow 25x-75y=-600 \\ & \Rightarrow x-3y=-24 \\ \end{align}
Now, from the simplification of first equation we know that $2x=3y$ therefore using this in last equation we get
\begin{align} & x-2x=-24 \\ & \Rightarrow x=24 \\ \end{align}
And putting this in the first equation we get $y=16$ .